z=0為2級極點
∫sinz/z²dz=2πiRes[sinz/z²,0]=2πi*[1/(2-1)!]lim[z->0]{d²[(z²)*(sinz/z²)]/dz²}=2πi*(-sin0)=0
z=0,z=-1為單極點
∫cosz/[z(z+1)]dz=2πi{Res[cosz/[z(z+1)],0] + Res[cosz/[z(z+1)],-1]}=2πi(-1-sin1-cos1)
x=±i為單極點,但上半平面只有x=i
∫[∞→0](cos^2)x/(1+x^2)dx=(-1/2)∫[-∞→∞](cos2x+1)/2(1+x²)dx
=(-1/2)*[∫[-∞→∞] cos2x/2(1+x²)dx+∫[-∞→∞] 1/2(1+x²)dx]
=(-1/2)*[Re{∫[-∞→∞] e^(2ix)/2(1+x²)dx} + (1/2)(arctan∞-arctan(-∞))]
=(-1/2)*[Re{2πiRes[e^(2ix)/2(1+x²),i]}-π/4
=(-1/2)*[Re{2πi*5/8e²}]-π/4
=0-π/4
=-π/4